Some sulfuric acid is spilled on a lab bench. It can be neutralized by sprinkling sodium bicarbonate on it and then mopping up the resultant solution.
The sodium bicarbonate reacts with sulfuric acid as follows.
2 NaHCO3(s) + H2SO4(aq) --> Na2SO4(aq) + 2 H2O(l) + 2 CO2(g)
Sodium bicarbonate is added until the fizzing due to the formation of CO2(g) stops. If 34 mL of 6.0 M H2SO4 was spilled, what is the minimum mass of NaHCO3 that must be added to the spill to neutralize the acid?
i got 5.7 but it said:
"Your response differs from the correct answer by more than 10%. Double check your calculations."
what am i doing wrong?
answer in book is 27g
i don't think 34 is correct because that is way more than 10% of my first answer of 5.7
I worked the same problem for another student a day or so ago and came up with approximately 34g as I remember. And if you will look at the response to your 5.7 value, it says "Your response differs from the correct answer by more than 10%. Double check your calculations." The bold faced print is mine. So why can't 34 be correct? It differs, not by 10% which you imply it should, but by MORE THAN 10%, which it does.
57.13
Well, it seems like you're getting pretty close with your answer but just missing the mark. Let's go through the calculations again and see if we can figure out what went wrong!
First, let's calculate the number of moles of H2SO4 that was spilled:
moles H2SO4 = volume (L) x concentration (M)
moles H2SO4 = 0.034 L x 6.0 M
moles H2SO4 = 0.204 mol
According to the balanced equation, we have a 1:2 ratio between moles of H2SO4 and moles of NaHCO3. This means we'll need twice as many moles of NaHCO3 to neutralize the acid.
moles NaHCO3 = 2 x moles H2SO4
moles NaHCO3 = 2 x 0.204 mol
moles NaHCO3 = 0.408 mol
Finally, let's calculate the mass of NaHCO3 needed using its molar mass:
mass NaHCO3 = moles NaHCO3 x molar mass NaHCO3
mass NaHCO3 = 0.408 mol x (22.99 g/mol + 1.01 g/mol + 12.01 g/mol + 3 x 16.00 g/mol)
mass NaHCO3 = 0.408 mol x 84.01 g/mol
mass NaHCO3 = 34.996 g
So the minimum mass of NaHCO3 that must be added to neutralize the acid is approximately 34.996 grams. Keep in mind that we generally round our answer to the appropriate number of significant figures, but in this case, the given value of 34 mL H2SO4 determined the precision of our final answer.
I hope that helps! Keep up the great work!
To find the minimum mass of NaHCO3 needed to neutralize the acid, we can use stoichiometry. Here's how you can calculate it:
1. Determine the number of moles of H2SO4 spilled:
Number of moles = concentration × volume
Number of moles = 6.0 M × 0.034 L
Number of moles = 0.204
2. According to the balanced equation, 2 moles of NaHCO3 are required to react with 1 mole of H2SO4. Therefore, we can use the mole ratio to calculate the moles of NaHCO3 needed:
Number of moles of NaHCO3 = (2/1) × 0.204
Number of moles of NaHCO3 = 0.408
3. Finally, calculate the mass of NaHCO3 required using the molar mass:
Mass = moles × molar mass
Mass = 0.408 × 84.01 g/mol
Mass ≈ 34.315 g
Hence, the minimum mass of NaHCO3 that must be added to neutralize the acid is approximately 34.315 grams.
If you initially got 5.7 grams, it seems that you made an error in your calculation. Double-check the steps to ensure correct calculation, including using the correct molar mass.
The equation tells you that 2 moles of bicarbonate react with each mole of H2SO4
number of moles of H2SO4
6.0 mole L^-1 x 0.034 L= 0.204 mole
so 0.408 moles of bicarbonate are needed.
The molar mass of NaHCO3 is
84 g mole^-1
so mass of NaHCO3
0.408 mole x 84 g mole^-1 =34 g
2 sig figs appropriate in the answer as both starting values are to 2 sig figs.